New Image Question

artisticre

I have a page that I am creating. It is a photo gallery. I have this working. What I now want to do is when someone clicks on the image, I want to have a page dynamically created based on the image name. Can someone help me out?

thanks

seby

are you storing the image name in the db?

the way i handle my images is when they are uploaded a record gets inserted into the db, then i call the images like image.php?id=1 and it will display image with id 1 ..

artisticre

I am storing the image name in the database with the actual image in a folder on the server.

select * from $dbtable ORDER BY image

<a href=\"$linkpath".$result[image]."$shtml"."\"><img border=1 height=123 width=95 alt=\"".$result[shirtname]."\" src=\"".$path.$result[image].$type."\"></a> $result[shirtname]

When they click on the image, I would like to have a page created based on the imagename and pull things like size & price. I know how to pull the information, I just don't know how to do it based on the image they click

seby

as i've said.. If your images name are stored in the db and each row has an ID all you have to do is fetch the image based GET.

example:

SELECT * FROM images where id = '$_GET[id]'

Then you can do page.php?id=AN_ID and it will grab that image....

If you are doing it by image name then its no different:

SELECT * FROM images where image_name = '$_GET[image]'

Then you can do page.php?image=mypic.jpg and it will grab that image....

artisticre

I want to create a template page once. Then, when the image is clicked, have this template create a page based on the image id. This will show text data. I understand how to get the image to show, but how do I do it to get the text information. Please bear with me, I am still learning this stuff.

thanks

artisticre

The following code creates the photo gallery with 12 images per page and then paginates to the next page. What I would like to do is when a user clicks on one of the images, dynamically create a page so I don't have to create a page for each image since I have at least 100 images. When this page is created, I would like to have it pull some information from the database. I know how to pull information from the database, I am not sure how to do it when clicking on a link and how to do it based on that link.

thanks

$server = "**";
$userid = "";
$pass = "";
$database = "**";
$dbtable = "tshirts";
$limit = 12;

$con = mysql_connect("$server","$userid","$pass") or die ("Huh? What Server");
$db = mysql_select_db("$database",$con) or die("I said WHAT database");
if (empty($offset) || $offset < 0) {
$offset=0;
}
if (empty($index)) $index=0;
$getrows = mysql_query("select from $dbtable", $con);
$numrows=mysql_num_rows($getrows);
$path ="/NewCindys/Images/TShirts/";
$shtml =".shtml";
$linkpath ="/TShirts/";
$query = mysql_query ("select from $dbtable where id ='$_GET[id]' limit $offset,$limit", $con);

if ($numrows <= $limit) {
}
else {
if ($offset!=0) {
$prevoffset=$offset-$limit;
echo "<a onMouseOver=\"window.status='Previous $limit Results'; return true\"; href=\"$PHP_SELF?offset=$prevoffset&index=$prevoffset\">[Previous]</a>";
}
else echo "[Previous] ";
$pages = intval($numrows/$limit);
if ($numrows%$limit) {
$pages++;
}
for ($i=1;$i<=$pages;$i++) {
if (($offset/$limit) == ($i-1)) {
echo " $i ";
} else {
$newoffset=$limit*($i-1);
echo " <a onMouseOver=\"window.status='Page $i Results'; return true\"; href=\"$PHP_SELF?offset=$newoffset&index=$newoffset\">$i</a> \n";
}

}
if (!((($offset/$limit)+1)==$pages) && $pages!=1) {
$newoffset=$offset+$limit;
echo " <a onMouseOver=\"window.status='Next $limit Results'; return true\"; href=\"$PHP_SELF?offset=$newoffset&index=$newoffset\">[Next]</a>\n";
} else echo "[Next]";
}
echo "<table align=center border=0><tr>";
$num=1;
while ($result=mysql_fetch_array($query)){
$num = ($num < 5 ? $num : 1);
$index++; / Increment the line index by 1 /

echo "<td align=center class='shirts'><a href=\"$linkpath".$result[image]."$shtml"."\"><img border=1 height=123 width=95 alt=\"".$result[shirtname]."\" src=\"".$path.$result[id].$type."\"></a> $result[shirtname]</td>";
if ($num==4 && $index!=$numrows)
echo '</tr><tr>';

$num++;
}
echo ("</tr></table>");
mysql_close($con);
?>

Kinsbane

What you're probably gonna hafta do is when a new image is added to the db, add a script to it that writes out a nicely formatted table with all of your variables in the right places ready to output - IIRC, the function is fopen() or something like that - basically what could happen is you have 1 page that will fill in the info for any page/image - say like, imageview.php - then, in your grid of 12 images, make sure your links say something like, 'imageview.php?id=001' , where 001 would be the unique ID of the image. Next, in your imageview.php page, write out a nicely formatted table to display the image - this time using the $_GET['id'] function to return the image to display - or you could even retrieve the full url of the image and echo it into an img src="" tag. Just some assumptions 🙂